Question

If $p{\lambda }^4+q{\lambda }^3+r{\lambda }^2+s\lambda +t=\left|\begin{array}{ccc}{\lambda }^2+3& \lambda -1& \lambda +3\\ {\lambda }^2+1& 2-\lambda & \lambda -3\\ {\lambda }^2-3& \lambda +4& 3\lambda \end{array}\right|$, then $p$ is equal to

• A. $-5$
• B. $-4$
• C. $-3$
• D. $-2$

Answer 1

The correct option is B $-4$

$p{\lambda }^4+q{\lambda }^3+r{\lambda }^2+s\lambda +t=\left|\begin{array}{ccc}{\lambda }^2+3& \lambda -1& \lambda +3\\ {\lambda }^2+1& 2-\lambda & \lambda -3\\ {\lambda }^2-3& \lambda +4& 3\lambda \end{array}\right|$
Consider, $\left|\begin{array}{ccc}{\lambda }^2+3& \lambda -1& \lambda +3\\ {\lambda }^2+1& 2-\lambda & \lambda -3\\ {\lambda }^2-3& \lambda +4& 3\lambda \end{array}\right|$
$R_1\to R_1-R_3,R_2\to R_2-R_3$
$=\left|\begin{array}{ccc}6& -5& -2\lambda +3\\ 4& -2\lambda -2& -2\lambda -3\\ {\lambda }^2-3& \lambda +4& 3\lambda \end{array}\right|$
$\Rightarrow -4{\lambda }^4+12{\lambda }^3+{\lambda }^2+34\lambda +57$
Comparing with the LHS of given equation,
$p=-4$